prove that all pure states in a commutative C* algebra are multiplicative linear functionals

Question

I am trying to prove this , but can not see it clearly. it was given as some sort of converse of the fact that all multiplicative linear functionals are pure states

Answer 1

There might be an algebraic way, but the canonical way is to see your algebra as $C_0(X)$ and then show that pure states are point evaluations, which are clearly multiplicative.

Answer 2

Let $A$ be commutative and $\tau:A\to \Bbb C $ be a pure state and $(\pi,H, \xi)$ be its corresponding representation.

Commutativity of $A$ implies that $\pi(A)\subset\pi (A)'$. Also $\tau$ is pure, which consequences that $\pi(A)'=\Bbb C1$.

Now for $a,b \in A$, we have $$\tau(ab)= \langle \pi(a)\pi(b)\xi,\xi\rangle = \langle \pi(a)\xi, \xi\rangle \langle \pi(b)\xi,\xi\rangle = \tau(a)\tau(b)$$